204. Count Primes (java)

最新推荐文章于 2021-02-13 14:09:33 发布

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本文介绍了一种高效计算小于非负整数n的所有素数数量的方法。通过使用布尔数组筛除不符合条件的数字，显著提高了算法的时间效率。

204. Count Primes

Description Submissions Solutions

Total Accepted: 106085
Total Submissions: 403293
Difficulty: Easy
Contributor: LeetCode

Description:

Count the number of prime numbers less than a non-negative number, n.

Credits:
Special thanks to @mithmatt for adding this problem and creating all test cases.

Hint:

Let's start with a isPrime function. To determine if a number is prime, we need to check if it is not divisible by any number less than n. The runtime complexity of isPrime function would be O(n) and hence counting the total prime numbers up to n would be O(n²). Could we do better?

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题目链接：https://leetcode.com/problems/count-primes/#/description

解题思路：求小于n的所有素数的个数。这题对时间效率要求较高，一般O(n^2)的方法不能用，用布尔数组来筛掉不符合条件的数字，该数字就不再进入下一层循环进行判断了。

public class Solution {
    public int countPrimes(int n) {
        boolean[] vis= new boolean[n];
        Arrays.fill(vis,false);
        int cnt=0;
        for(int i=2;i<n;i++){
            if(vis[i]==false){
                cnt++;
                for(int j=2;j*i<n;j++){
                    vis[i*j]=true;              
                }
            }
        }
        return cnt;
    }
}